Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
![**Dilution Calculation**
---
**Problem Statement:**
How many mL of 1.5 M KMnO₄ are required to make 1.50 L of 0.005 M?
*Equation Box:* [____________]
---
**Explanation:**
To solve this dilution problem, you can use the dilution equation, which is:
\[ C_1 V_1 = C_2 V_2 \]
where
- \( C_1 \) is the concentration of the stock solution (1.5 M),
- \( V_1 \) is the volume of the stock solution required,
- \( C_2 \) is the concentration of the dilute solution (0.005 M), and
- \( V_2 \) is the volume of the dilute solution (1.50 L).
Rearrange to solve for \( V_1 \):
\[ V_1 = \dfrac{C_2 \times V_2}{C_1} \]
Substitute the given values:
\[ V_1 = \dfrac{0.005 \times 1.50}{1.5} \]
Calculate \( V_1 \):
\[ V_1 = \dfrac{0.0075}{1.5} = 0.005 \text{ L} \]
Convert liters to milliliters:
\[ V_1 = 0.005 \times 1000 = 5 \text{ mL} \]
Therefore, 5 mL of 1.5 M KMnO₄ is required to make 1.50 L of 0.005 M solution.
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19b0aed3-745e-4879-8577-ea35399669d0%2F52d590a7-c4a8-4f85-9fdf-c98f8a87dd63%2Fv9krzjp_processed.png&w=3840&q=75)
Transcribed Image Text:**Dilution Calculation**
---
**Problem Statement:**
How many mL of 1.5 M KMnO₄ are required to make 1.50 L of 0.005 M?
*Equation Box:* [____________]
---
**Explanation:**
To solve this dilution problem, you can use the dilution equation, which is:
\[ C_1 V_1 = C_2 V_2 \]
where
- \( C_1 \) is the concentration of the stock solution (1.5 M),
- \( V_1 \) is the volume of the stock solution required,
- \( C_2 \) is the concentration of the dilute solution (0.005 M), and
- \( V_2 \) is the volume of the dilute solution (1.50 L).
Rearrange to solve for \( V_1 \):
\[ V_1 = \dfrac{C_2 \times V_2}{C_1} \]
Substitute the given values:
\[ V_1 = \dfrac{0.005 \times 1.50}{1.5} \]
Calculate \( V_1 \):
\[ V_1 = \dfrac{0.0075}{1.5} = 0.005 \text{ L} \]
Convert liters to milliliters:
\[ V_1 = 0.005 \times 1000 = 5 \text{ mL} \]
Therefore, 5 mL of 1.5 M KMnO₄ is required to make 1.50 L of 0.005 M solution.
---
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